Question

The diagonal of a square is changing at the rate of $0.5$ cm/sec. Then the rate of change of area when the area is $400$ cm$^{2}$ is equal to}

• A. $20\sqrt{2}$ cm$^{2}$/sec
• B. $10\sqrt{2}$ cm$^{2}$/sec
• C. $\frac{1}{10\sqrt{2$ cm$^{2}$/sec
• D. $\frac{10}{\sqrt{2$ cm$^{2}$/sec

Hint:

For related rates problems, first identify all variables and their relationships (e.g., geometric formulas). Differentiate the primary equation with respect to time. Then, substitute known rates and specific values to find the unknown rate. Always pay attention to units.

Answer 1

The Correct Option is C

Step 1: Let $s$ be the side length of the square and $x$ be the length of its diagonal. By Pythagorean theorem, $s^{2} + s^{2} = x^{2}$, so $2s^{2} = x^{2}$, which means $s = \frac{x}{\sqrt{2$.\
Step 2: The area of the square, $A$, is $s^{2}$. Substitute $s = \frac{x}{\sqrt{2$ into the area formula:\ \[ A = \left(\frac{x}{\sqrt{2\right)^{2} \] \[ A = \frac{x^{2{2} \]\
Step 3: We are given that the diagonal is changing at the rate of $0.5$ cm/sec, so $\frac{dx}{dt} = 0.5$ cm/sec. We need to find $\frac{dA}{dt}$ when $A = 400$ cm$^{2}$.\
Step 4: Differentiate the area formula $A = \frac{x^{2{2}$ with respect to time $t$.\ \[ \frac{dA}{dt} = \frac{1}{2} \cdot 2x \frac{dx}{dt} \] \[ \frac{dA}{dt} = x \frac{dx}{dt} \]\

Step 5: Find the value of $x$ when $A = 400$ cm$^{2}$.\ \[ 400 = \frac{x^{2{2} \] \[ x^{2} = 800 \] \[ x = \sqrt{800} \] \[ x = \sqrt{400 \times 2} \] \[ x = 20\sqrt{2} \text{ cm} \]\
Step 6: Substitute the values of $x$ and $\frac{dx}{dt}$ into the expression for $\frac{dA}{dt}$.\ \[ \frac{dA}{dt} = (20\sqrt{2}) (0.5) \] \[ \frac{dA}{dt} = 10\sqrt{2} \text{ cm}^{2}\text{/sec} \]